Question

A segment of a circle has a 120 arc and a chord of 8in. Find the area of the segment

Answer 1

`\text{Area of segment =}13.102 in^2`

Explanation:

Given a segment of a circle has a 120 arc and a chord of 8 in. We have to find the area of segment.

As, radius from the centre perpendicularly bisect the chord ∴ 120 degree arc is also bisected.

Hence, we get a right triangle with the radius of circle the hypotenuse and 4 opposite side and the 60 degree angle i.e

In ΔOAB

AB=4 in

∠AOB=60°

`\sin60=(AB)/(OB)`

⇒
`Radius=OB=4 * \csc(60) = 4 * (2)/(\sqrt3) = (8)/(sqrt3)in`

`\text{Then},\text{ the area of the sector is }=(1)/(2)r^2\theta=(1)/(2)((8)/(\sqrt3))^2 120((\pi)/(180))=(64\pi)/(9)in^2`

Now, we have to find the area of triangle ODB

`Base=DB=8 in`

`Height=OA=(AB)/(\tan60)=(4)/(\sqrt3)in`

`\text{Area of triangle ODB=}(1)/(2)* DB* OA=(1)/(2)*8* (4)/(\sqrt3)=(16)/(\sqrt3)in^2`

`\text{Area of segment =} (64\pi)/(9) - (16)/(\sqrt3)=13.102 in^2`

Answer 2

We calculate as follows:

The radius must be 8/sqrt(3)
So area of entire circle is pi * 64/3
So area of sector is 64pi/9
area of segment = 64pi/9 - 16sqrt(3)/3 = 13.102 in^2

Hope this answers the question. Have a nice day.